Tính :
( 2a + b - 5 ) x ( 2a - b + 5 )
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DA
0
24 tháng 8 2021
Bài 2:
a) \(\left(x+5\right)^2=x^2+10x+25\)
b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)
c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)
d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)
e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)
f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
Bài 1:
$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$
$=4a.2b=8ab$
$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$
$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$
$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$
$=m^2+2mn+n^2=(m+n)^2$
17 tháng 4 2023
a: \(=-4x^2+20x+2x-10=-4x^2+22x-10\)
b: =x^2-9
c: =x^3+27
d: \(=-2x^2-6x+x+3=-2x^2-5x+3\)
e: =8a^3+1
f: =(3-x)(x+1)(x+2)
=(3-x)(x^2+3x+2)
=3x^2+9x+6-x^3-3x^2-2x
=-x^3+7x+6
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 4 2022
a.
\(tana=\dfrac{sina}{cosa}=\dfrac{1}{15}\Rightarrow sina=\dfrac{cosa}{15}\)
\(\Rightarrow sin2a=2sina.cosa=\dfrac{2cosa}{15}.cosa=\dfrac{2}{15}cos^2a=\dfrac{2}{15}.\dfrac{1}{1+tan^2a}=\dfrac{2}{15}.\dfrac{1}{1+\dfrac{1}{15^2}}=\dfrac{15}{113}\)
b.
\(5^2=\left(3sina+4cosa\right)^2\le\left(3^2+4^2\right)\left(sin^2+cos^2a\right)=25\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{sina}{3}=\dfrac{cosa}{4}\\3sina+4cosa=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}sina=\dfrac{3}{5}\\cosa=\dfrac{4}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 4 2022
c.
\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
\(\Leftrightarrow\dfrac{cos^2a}{sin^2a}+\dfrac{sin^2a}{cos^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
\(\)\(\Leftrightarrow\dfrac{sin^4a+cos^4a}{sin^2a.cos^2a}+\dfrac{sin^2a+cos^2a}{sin^2a.cos^2a}=7\)
\(\Leftrightarrow\dfrac{\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a}{sin^2a.cos^2a}+\dfrac{1}{sin^2a.cos^2a}=7\)
\(\Leftrightarrow\dfrac{2}{sin^2a.cos^2a}=9\)
\(\Leftrightarrow\dfrac{8}{\left(2sina.cosa\right)^2}=9\)
\(\Leftrightarrow\dfrac{8}{sin^22a}=9\)
\(\Leftrightarrow sin^22a=\dfrac{8}{9}\)
19 tháng 9 2021
\(\left(2a+b-5\right)\left(2a-b+5\right)\)
\(=\left[2a+\left(b-5\right)\right]\left[2a-\left(b-5\right)\right]\)
\(=4a^2-\left(b-5\right)^2\)
\(=4a^2-\left(b^2-10b+25\right)\)
\(=4a^2-b^2+10b-25\)
19 tháng 9 2021
\(\left(2a+b-5\right)\left(2a-b+5\right)\)
\(=\left(2a\right)^2-\left(b-5\right)^2\)
\(=4a^2-\left(b-5\right)^2\)
\(\left(2a+b-5\right)\left(2a-b+5\right)=\left[2a+\left(b-5\right)\right]\left[2a-\left(b-5\right)\right]=4a^2-\left(b-5\right)^2hoặc\left(2a+b-5\right)\left(2a-b+5\right)=4a^2-2ab+10a+2ab-b^2+5b-10a+5b-25=4a^2-b^2+10b-25=4a^2-\left(b-5\right)^2\)
(2a+b-5).(2a-b+5)
=2a(2a-b+5)+b(2a-b+5)-5(2a-b+5)
=4a\(^2\) -2ab+10a+2ab-b\(^2\)+5b-10a+5b-25
=4a\(^{2^{ }}\)-25-b\(^2\)